\(\int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx\) [174]

Optimal result

Integrand size = 13, antiderivative size = 143 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[Out]

-3*b^2/(b*x-arccoth(tanh(b*x+a)))^3/arccoth(tanh(b*x+a))+3/2*b/x/(b*x-arccoth(tanh(b*x+a)))^2/arccoth(tanh(b*x
+a))+1/2/x^2/(b*x-arccoth(tanh(b*x+a)))/arccoth(tanh(b*x+a))+3*b^2*ln(x)/(b*x-arccoth(tanh(b*x+a)))^4-3*b^2*ln
(arccoth(tanh(b*x+a)))/(b*x-arccoth(tanh(b*x+a)))^4

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {2202, 2194, 2191, 2188, 29} \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))} \]

[In]

Int[1/(x^3*ArcCoth[Tanh[a + b*x]]^2),x]

[Out]

(-3*b^2)/((b*x - ArcCoth[Tanh[a + b*x]])^3*ArcCoth[Tanh[a + b*x]]) + (3*b)/(2*x*(b*x - ArcCoth[Tanh[a + b*x]])
^2*ArcCoth[Tanh[a + b*x]]) + 1/(2*x^2*(b*x - ArcCoth[Tanh[a + b*x]])*ArcCoth[Tanh[a + b*x]]) + (3*b^2*Log[x])/
(b*x - ArcCoth[Tanh[a + b*x]])^4 - (3*b^2*Log[ArcCoth[Tanh[a + b*x]]])/(b*x - ArcCoth[Tanh[a + b*x]])^4

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2188

Int[(u_)^(m_.), x_Symbol] :> With[{c = Simplify[D[u, x]]}, Dist[1/c, Subst[Int[x^m, x], x, u], x]] /; FreeQ[m,
x] && PiecewiseLinearQ[u, x]

Rule 2191

Int[1/((u_)*(v_)), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Dist[b/(b*u - a*v), Int[1
/v, x], x] - Dist[a/(b*u - a*v), Int[1/u, x], x] /; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x]

Rule 2194

Int[(v_)^(n_)/(u_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[v^(n + 1)/((n + 1)*
(b*u - a*v)), x] - Dist[a*((n + 1)/((n + 1)*(b*u - a*v))), Int[v^(n + 1)/u, x], x] /; NeQ[b*u - a*v, 0]] /; Pi
ecewiseLinearQ[u, v, x] && LtQ[n, -1]

Rule 2202

Int[(u_)^(m_)*(v_)^(n_), x_Symbol] :> With[{a = Simplify[D[u, x]], b = Simplify[D[v, x]]}, Simp[(-u^(m + 1))*(
v^(n + 1)/((m + 1)*(b*u - a*v))), x] + Dist[b*((m + n + 2)/((m + 1)*(b*u - a*v))), Int[u^(m + 1)*v^n, x], x] /
; NeQ[b*u - a*v, 0]] /; PiecewiseLinearQ[u, v, x] && NeQ[m + n + 2, 0] && LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {(3 b) \int \frac {1}{x^2 \coth ^{-1}(\tanh (a+b x))^2} \, dx}{2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right )} \\ & = \frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 b^2\right ) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))^2} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )} \\ & = -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}-\frac {\left (3 b^2\right ) \int \frac {1}{x \coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {\left (3 b^2\right ) \int \frac {1}{x} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2}-\frac {\left (3 b^3\right ) \int \frac {1}{\coth ^{-1}(\tanh (a+b x))} \, dx}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {\left (3 b^2\right ) \text {Subst}\left (\int \frac {1}{x} \, dx,x,\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^2} \\ & = -\frac {3 b^2}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^3 \coth ^{-1}(\tanh (a+b x))}+\frac {3 b}{2 x \left (b x-\coth ^{-1}(\tanh (a+b x))\right )^2 \coth ^{-1}(\tanh (a+b x))}+\frac {1}{2 x^2 \left (b x-\coth ^{-1}(\tanh (a+b x))\right ) \coth ^{-1}(\tanh (a+b x))}+\frac {3 b^2 \log (x)}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4}-\frac {3 b^2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )}{\left (b x-\coth ^{-1}(\tanh (a+b x))\right )^4} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.64 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {2 b^3 x^3-6 b x \coth ^{-1}(\tanh (a+b x))^2+\coth ^{-1}(\tanh (a+b x))^3-3 b^2 x^2 \coth ^{-1}(\tanh (a+b x)) \left (-1+2 \log (x)-2 \log \left (\coth ^{-1}(\tanh (a+b x))\right )\right )}{2 x^2 \coth ^{-1}(\tanh (a+b x)) \left (-b x+\coth ^{-1}(\tanh (a+b x))\right )^4} \]

[In]

Integrate[1/(x^3*ArcCoth[Tanh[a + b*x]]^2),x]

[Out]

-1/2*(2*b^3*x^3 - 6*b*x*ArcCoth[Tanh[a + b*x]]^2 + ArcCoth[Tanh[a + b*x]]^3 - 3*b^2*x^2*ArcCoth[Tanh[a + b*x]]
*(-1 + 2*Log[x] - 2*Log[ArcCoth[Tanh[a + b*x]]]))/(x^2*ArcCoth[Tanh[a + b*x]]*(-(b*x) + ArcCoth[Tanh[a + b*x]]
)^4)

Maple [F(-1)]

Timed out.

\[\int \frac {1}{x^{3} \operatorname {arccoth}\left (\tanh \left (b x +a \right )\right )^{2}}d x\]

[In]

int(1/x^3/arccoth(tanh(b*x+a))^2,x)

[Out]

int(1/x^3/arccoth(tanh(b*x+a))^2,x)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.27 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.71 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {2 \, {\left (48 \, a b^{2} x^{2} + 24 \, a^{2} b x + i \, \pi ^{3} - 6 \, \pi ^{2} {\left (b x - a\right )} - 8 \, a^{3} + 12 i \, \pi {\left (2 \, b^{2} x^{2} + 2 \, a b x - a^{2}\right )} - 24 \, {\left (2 \, b^{3} x^{3} + i \, \pi b^{2} x^{2} + 2 \, a b^{2} x^{2}\right )} \log \left (i \, \pi + 2 \, b x + 2 \, a\right ) + 24 \, {\left (2 \, b^{3} x^{3} + i \, \pi b^{2} x^{2} + 2 \, a b^{2} x^{2}\right )} \log \left (x\right )\right )}}{32 \, a^{4} b x^{3} + i \, \pi ^{5} x^{2} + 32 \, a^{5} x^{2} + 2 \, \pi ^{4} {\left (b x^{3} + 5 \, a x^{2}\right )} - 8 i \, \pi ^{3} {\left (2 \, a b x^{3} + 5 \, a^{2} x^{2}\right )} - 16 \, \pi ^{2} {\left (3 \, a^{2} b x^{3} + 5 \, a^{3} x^{2}\right )} + 16 i \, \pi {\left (4 \, a^{3} b x^{3} + 5 \, a^{4} x^{2}\right )}} \]

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="fricas")

[Out]

2*(48*a*b^2*x^2 + 24*a^2*b*x + I*pi^3 - 6*pi^2*(b*x - a) - 8*a^3 + 12*I*pi*(2*b^2*x^2 + 2*a*b*x - a^2) - 24*(2
*b^3*x^3 + I*pi*b^2*x^2 + 2*a*b^2*x^2)*log(I*pi + 2*b*x + 2*a) + 24*(2*b^3*x^3 + I*pi*b^2*x^2 + 2*a*b^2*x^2)*l
og(x))/(32*a^4*b*x^3 + I*pi^5*x^2 + 32*a^5*x^2 + 2*pi^4*(b*x^3 + 5*a*x^2) - 8*I*pi^3*(2*a*b*x^3 + 5*a^2*x^2) -
16*pi^2*(3*a^2*b*x^3 + 5*a^3*x^2) + 16*I*pi*(4*a^3*b*x^3 + 5*a^4*x^2))

Sympy [F]

\[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\int \frac {1}{x^{3} \operatorname {acoth}^{2}{\left (\tanh {\left (a + b x \right )} \right )}}\, dx \]

[In]

integrate(1/x**3/acoth(tanh(b*x+a))**2,x)

[Out]

Integral(1/(x**3*acoth(tanh(a + b*x))**2), x)

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.53 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.33 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {48 \, b^{2} \log \left (-i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {48 \, b^{2} \log \left (x\right )}{\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {2 \, {\left (24 \, b^{2} x^{2} + \pi ^{2} + 4 i \, \pi a - 4 \, a^{2} - 6 \, {\left (i \, \pi b - 2 \, a b\right )} x\right )}}{2 \, {\left (i \, \pi ^{3} b - 6 \, \pi ^{2} a b - 12 i \, \pi a^{2} b + 8 \, a^{3} b\right )} x^{3} + {\left (\pi ^{4} + 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} - 32 i \, \pi a^{3} + 16 \, a^{4}\right )} x^{2}} \]

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="maxima")

[Out]

-48*b^2*log(-I*pi + 2*b*x + 2*a)/(pi^4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) + 48*b^2*log(x)/(pi^
4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi*a^3 + 16*a^4) + 2*(24*b^2*x^2 + pi^2 + 4*I*pi*a - 4*a^2 - 6*(I*pi*b - 2
*a*b)*x)/(2*(I*pi^3*b - 6*pi^2*a*b - 12*I*pi*a^2*b + 8*a^3*b)*x^3 + (pi^4 + 8*I*pi^3*a - 24*pi^2*a^2 - 32*I*pi
*a^3 + 16*a^4)*x^2)

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.29 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.42 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=-\frac {48 \, b^{2} \log \left (i \, \pi + 2 \, b x + 2 \, a\right )}{\pi ^{4} - 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} + 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {48 \, b^{2} \log \left (x\right )}{\pi ^{4} - 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} + 32 i \, \pi a^{3} + 16 \, a^{4}} + \frac {16 \, b^{2}}{-2 i \, \pi ^{3} b x - 12 \, \pi ^{2} a b x + 24 i \, \pi a^{2} b x + 16 \, a^{3} b x + \pi ^{4} - 8 i \, \pi ^{3} a - 24 \, \pi ^{2} a^{2} + 32 i \, \pi a^{3} + 16 \, a^{4}} - \frac {4 \, {\left (i \, \pi - 8 \, b x + 2 \, a\right )}}{-2 i \, \pi ^{3} x^{2} - 12 \, \pi ^{2} a x^{2} + 24 i \, \pi a^{2} x^{2} + 16 \, a^{3} x^{2}} \]

[In]

integrate(1/x^3/arccoth(tanh(b*x+a))^2,x, algorithm="giac")

[Out]

-48*b^2*log(I*pi + 2*b*x + 2*a)/(pi^4 - 8*I*pi^3*a - 24*pi^2*a^2 + 32*I*pi*a^3 + 16*a^4) + 48*b^2*log(x)/(pi^4
- 8*I*pi^3*a - 24*pi^2*a^2 + 32*I*pi*a^3 + 16*a^4) + 16*b^2/(-2*I*pi^3*b*x - 12*pi^2*a*b*x + 24*I*pi*a^2*b*x
+ 16*a^3*b*x + pi^4 - 8*I*pi^3*a - 24*pi^2*a^2 + 32*I*pi*a^3 + 16*a^4) - 4*(I*pi - 8*b*x + 2*a)/(-2*I*pi^3*x^2
- 12*pi^2*a*x^2 + 24*I*pi*a^2*x^2 + 16*a^3*x^2)

Mupad [B] (verification not implemented)

Time = 7.85 (sec) , antiderivative size = 689, normalized size of antiderivative = 4.82 \[ \int \frac {1}{x^3 \coth ^{-1}(\tanh (a+b x))^2} \, dx=\frac {2\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3-2\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^3-6\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2+6\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-32\,b^3\,x^3+24\,b\,x\,{\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2+24\,b\,x\,{\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )}^2-24\,b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+24\,b^2\,x^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-b^2\,x^2\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,96{}\mathrm {i}-48\,b\,x\,\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+b^2\,x^2\,\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,\mathrm {atan}\left (\frac {\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}-\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\,1{}\mathrm {i}+b\,x\,2{}\mathrm {i}}{\ln \left (-\frac {2}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {2\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x}\right )\,96{}\mathrm {i}}{x^2\,\left (\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )\right )\,{\left (\ln \left (-\frac {1}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )-\ln \left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}}{{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}-1}\right )+2\,b\,x\right )}^4} \]

[In]

int(1/(x^3*acoth(tanh(a + b*x))^2),x)

[Out]

(2*log(-1/(exp(2*a)*exp(2*b*x) - 1))^3 - 2*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))^3 - 6*log((exp
(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-1/(exp(2*a)*exp(2*b*x) - 1))^2 + 6*log((exp(2*a)*exp(2*b*x))
/(exp(2*a)*exp(2*b*x) - 1))^2*log(-1/(exp(2*a)*exp(2*b*x) - 1)) - 32*b^3*x^3 + 24*b*x*log((exp(2*a)*exp(2*b*x)
)/(exp(2*a)*exp(2*b*x) - 1))^2 + 24*b*x*log(-1/(exp(2*a)*exp(2*b*x) - 1))^2 - 24*b^2*x^2*log((exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) - 1)) + 24*b^2*x^2*log(-1/(exp(2*a)*exp(2*b*x) - 1)) - b^2*x^2*log((exp(2*a)*exp(2*b
*x))/(exp(2*a)*exp(2*b*x) - 1))*atan((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*1i - log(-2/(exp(
2*a)*exp(2*b*x) - 1))*1i + b*x*2i)/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*
exp(2*b*x) - 1)) + 2*b*x))*96i - 48*b*x*log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1))*log(-1/(exp(2*a)*
exp(2*b*x) - 1)) + b^2*x^2*log(-1/(exp(2*a)*exp(2*b*x) - 1))*atan((log((2*exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2
*b*x) - 1))*1i - log(-2/(exp(2*a)*exp(2*b*x) - 1))*1i + b*x*2i)/(log(-2/(exp(2*a)*exp(2*b*x) - 1)) - log((2*ex
p(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x))*96i)/(x^2*(log((exp(2*a)*exp(2*b*x))/(exp(2*a)*exp(2*b
*x) - 1)) - log(-1/(exp(2*a)*exp(2*b*x) - 1)))*(log(-1/(exp(2*a)*exp(2*b*x) - 1)) - log((exp(2*a)*exp(2*b*x))/
(exp(2*a)*exp(2*b*x) - 1)) + 2*b*x)^4)